To prove that f^−1(G ∪ H) = f^−1(G) ∪ f^−1(H), we need to show that each set is a subset of the other.

Step 1: Show that f^−1(G ∪ H) is a subset of f^−1(G) ∪ f^−1(H)

Let x be an arbitrary element of f^−1(G ∪ H). This means that f(x) is in G ∪ H. Therefore, f(x) is in G or f(x) is in H. If f(x) is in G, then x is in f^−1(G). If f(x) is in H, then x is in f^−1(H). In either case, x is in f^−1(G) ∪ f^−1(H). Since x was an arbitrary element of f^−1(G ∪ H), we can conclude that f^−1(G ∪ H) is a subset of f^−1(G) ∪ f^−1(H).

Step 2: Show that f^−1(G) ∪ f^−1(H) is a subset of f^−1(G ∪ H)

Let x be an arbitrary element of f^−1(G) ∪ f^−1(H). This means that x is in f^−1(G) or x is in f^−1(H). If x is in f^−1(G), then f(x) is in G. If x is in f^−1(H), then f(x) is in H. In either case, f(x) is in G ∪ H, which means that x is in f^−1(G ∪ H). Since x was an arbitrary element of f^−1(G) ∪ f^−1(H), we can conclude that f^−1(G) ∪ f^−1(H) is a subset of f^−1(G ∪ H).

Since we have shown that each set is a subset of the other, we can conclude that f^−1(G ∪ H) = f^−1(G) ∪ f^−1(H).

Question

To prove that f^−1(G ∪ H) = f^−1(G) ∪ f^−1(H), we need to show that each set is a subset of the other.

Step 1: Show that f^−1(G ∪ H) is a subset of f^−1(G) ∪ f^−1(H)

Let x be an arbitrary element of f^−1(G ∪ H). This means that f(x) is in G ∪ H. Therefore, f(x) is in G or f(x) is in H. If f(x) is in G, then x is in f^−1(G). If f(x) is in H, then x is in f^−1(H). In either case, x is in f^−1(G) ∪ f^−1(H). Since x was an arbitrary element of f^−1(G ∪ H), we can conclude that f^−1(G ∪ H) is a subset of f^−1(G) ∪ f^−1(H).

Step 2: Show that f^−1(G) ∪ f^−1(H) is a subset of f^−1(G ∪ H)

Let x be an arbitrary element of f^−1(G) ∪ f^−1(H). This means that x is in f^−1(G) or x is in f^−1(H). If x is in f^−1(G), then f(x) is in G. If x is in f^−1(H), then f(x) is in H. In either case, f(x) is in G ∪ H, which means that x is in f^−1(G ∪ H). Since x was an arbitrary element of f^−1(G) ∪ f^−1(H), we can conclude that f^−1(G) ∪ f^−1(H) is a subset of f^−1(G ∪ H).

Since we have shown that each set is a subset of the other, we can conclude that f^−1(G ∪ H) = f^−1(G) ∪ f^−1(H).

Knowee AI · Accepted Answer

To prove that f^−1(G ∪ H) = f^−1(G) ∪ f^−1(H), we need to show that each set is a subset of the other.

Step 1: Show that f^−1(G ∪ H) is a subset of f^−1(G) ∪ f^−1(H)

Let x be an arbitrary element of f^−1(G ∪ H). This means that f(x) is in G ∪ H. Therefore, f(x) is in G or f(x) is in H. If f(x) is in G, then x is in f^−1(G). If f(x) is in H, then x is in f^−1(H). In either case, x is in f^−1(G) ∪ f^−1(H). Since x was an arbitrary element of f^−1(G ∪ H), we can conclude that f^−1(G ∪ H) is a subset of f^−1(G) ∪ f^−1(H).

Step 2: Show that f^−1(G) ∪ f^−1(H) is a subset of f^−1(G ∪ H)

Let x be an arbitrary element of f^−1(G) ∪ f^−1(H). This means that x is in f^−1(G) or x is in f^−1(H). If x is in f^−1(G), then f(x) is in G. If x is in f^−1(H), then f(x) is in H. In either case, f(x) is in G ∪ H, which means that x is in f^−1(G ∪ H). Since x was an arbitrary element of f^−1(G) ∪ f^−1(H), we can conclude that f^−1(G) ∪ f^−1(H) is a subset of f^−1(G ∪ H).

Since we have shown that each set is a subset of the other, we can conclude that f^−1(G ∪ H) = f^−1(G) ∪ f^−1(H).

Prove thatf −1(G ∪ H) = f −1(G) ∪ f −1(H)

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